Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
long long mpow(long long a, long long n, long long mod) {
  long long ret = 1;
  long long b = a;
  while (n) {
    if (n & 1) ret = (ret * b) % mod;
    b = (b * b) % mod;
    n >>= 1;
  }
  return (long long)ret;
}
using namespace std;
vector<int> graph[100000 + 5];
int depth[100000 + 5];
int timer = 0, n, maxa = 20;
int anc[100000 + 5][20];
vector<int> pep[100000 + 5][20];
vector<int> merg(vector<int>& x, vector<int>& y) {
  vector<int> ans;
  int i;
  for (i = 0; i < x.size(); i++) {
    ans.push_back(x[i]);
  }
  for (i = 0; i < y.size(); i++) {
    ans.push_back(y[i]);
  }
  sort(ans.begin(), ans.end());
  while (ans.size() > 10) ans.pop_back();
  return ans;
}
void dfs(int u, int par, int d) {
  anc[u][0] = par;
  depth[u] = d;
  for (int j = 1; j < maxa; j++) {
    anc[u][j] = anc[anc[u][j - 1]][j - 1];
    pep[u][j] = merg(pep[u][j - 1], pep[anc[u][j - 1]][j - 1]);
  }
  for (int j = 0; j < graph[u].size(); j++) {
    int to = graph[u][j];
    if (to != par) {
      dfs(to, u, d + 1);
    }
  }
}
int get_parent(int v, int k) {
  for (int i = 0; i < maxa; i++)
    if ((1 << i) & k) v = anc[v][i];
  return v;
}
int lca(int u, int v) {
  if (depth[u] < depth[v]) swap(u, v);
  u = get_parent(u, depth[u] - depth[v]);
  if (u == v) return u;
  for (int i = maxa - 1; i >= 0; i--)
    if (anc[u][i] != anc[v][i]) {
      u = anc[u][i];
      v = anc[v][i];
    }
  return anc[v][0];
}
vector<int> get(int v, int k) {
  vector<int> ans;
  for (int i = 0; i < maxa; i++) {
    if (k & (1 << i)) {
      ans = merg(ans, pep[v][i]);
      v = anc[v][i];
    }
  }
  return ans;
}
int main() {
  int n, i, j, m, q;
  scanf("%d%d%d", &n, &m, &q);
  for (j = 0; j < n - 1; j++) {
    int x, y;
    scanf("%d%d", &x, &y);
    graph[x].push_back(y);
    graph[y].push_back(x);
  }
  for (j = 1; j <= m; j++) {
    int x;
    scanf("%d", &x);
    if (pep[x][0].size() < 10) pep[x][0].push_back(j);
  }
  dfs(1, 0, 0);
  while (q--) {
    int u, v, a;
    scanf("%d%d%d", &u, &v, &a);
    int ans = lca(u, v);
    vector<int> x = get(u, depth[u] - depth[ans]);
    vector<int> y = get(v, depth[v] - depth[ans] + 1);
    x = merg(x, y);
    printf("%d ", min((int)x.size(), a));
    for (j = 0; j < min((int)x.size(), a); j++) printf("%d ", x[j]);
    printf("\n");
  }
  return 0;
}